# find the energy stored in coupled inductors

Discussion in 'Homework Help' started by rogerloh4.0, Nov 9, 2012.

1. ### rogerloh4.0 Thread Starter New Member

Sep 26, 2012
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Q: For the circuit in the figure, determine the energy stored in the coupled inductors at t = 1.5s.

A: 39.4J

Following is my approach on this problem:

1/8 F => -j4Ω
2H => j4Ω
1H => j2Ω

I determine I1 to be the current in the mesh on the primary side going clockwise, I2 to be the current in the mesh on the secondary side going clockwise.

40 = 4I1-j2I2
0 = (2+j2)I2-j2I1

I1 = (1-j)I2

(4-j6)I2 = 40

I2 = 5.547$\angle56.3^{\circ}$
I1 = 7.845$\angle11.3^{\circ}$

i2 = 5.547cos(2t + $\56.3^{\circ}$)
i1 = 7.845cos(2t + $\11.3^{\circ}$)

at t = 1.5s:

i1 = 7.845cos($\183.2^{\circ}$) = -7.833A
i2 = 5.547cos($\228.2^{\circ}$) = -3.697A

w = $\\frac{1}{2}*2*(-7.833)^{2}+\frac{1}{2}*(-3.697)^{2}+(-3.697)(-7.833)$ = 97.148J

Please tell me where have I got wrong, thanks!

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